Decinormal solution of NaCldeveloped an osmotic pressure of 4.6 atm at 300 K. Calculate its degree of dissociation. Share with your friends Share 10 Geetha answered this Calculated Osmotic pressure of a decinormal solution =CRT = 0.1 × 0.0821 × 300 = 2.46 atmObserved Osmotic pressure = 4.6 atmi = observed colligative propertyNormal colligative property = 4.62.46 = 1.87Degree of dissociation (α) = i -1α -1 =1.87 -12 -1 = 0.87% of dissociation = 100 × 0.87 = 87 % 52 View Full Answer
